Hello, Zak!

I assume you have a sketch . . .

A steel chimney is held in position by several guy wires.

Two such wires $\displaystyle AB$ and $\displaystyle AC$ where $\displaystyle B,\,C$ and the base of the chimney, $\displaystyle D$,

are on the same level in a straight line with $\displaystyle B$ and $\displaystyle C$ on the left of the chimney.

If $\displaystyle \text{angle }ABC = 45^o,\text{ angle }ACD = 67^o,\;BC = 32\text{ m,}$

find the height of the chimney, $\displaystyle AD.$ Code:

* A
* *|
* * |
* * |
* * | y
* * |
* * |
* 45° * 67° |
*---------------*-------*
B 32 C x D

Let $\displaystyle x = CD,\:y = AD.$

In right triangle $\displaystyle ADB:\;\tan45^o = \frac{y}{x+32}\quad\Rightarrow\quad1 = \frac{y}{x+32}\quad\Rightarrow\quad x \:= \:y - 32$ **[1]**

In right triangle $\displaystyle ADC:\;\tan67^o = \frac{y}{x}\quad\Rightarrow\quad x \,=\,\frac{y}{\tan67^o}$ **[2]**

Equate **[1]** and **[2]**: .$\displaystyle y - 32 \:=\:\frac{y}{\tan67^o}$

Solve for $\displaystyle y:\;\;y\tan67^o - 32\tan67^o \:=\:y\quad\Rightarrow\quad y\tan67^o - 32\tan67^o \:=\:y$

. . $\displaystyle y\tan67^o - y \:=\:32\tan67^o\quad \Rightarrow\quad y(\tan67^o - 1) \:=\:32\tan67^o$

Therefore: .$\displaystyle y \;=\;\frac{32\tan67^o}{\tan67^o - 1} \;=\;55.60138965 \;\approx\;\boxed{55.6\text{ m}}$